Re: [PHP-DEV] why does exit() print its argument? From: derick <email protected>
Date: 09/30/01

On Fri, 28 Sep 2001, Jim Winstead wrote:

> On Fri, Sep 28, 2001 at 11:14:04PM +0200, Zeev Suraski wrote:
> > At 22:17 28-09-01, Jason Greene wrote:
> > >Why does exit still set the exit status then?
> >
> > Ok, apparently it does (didn't recall that it does). I'm not sure what the
> > logic behind this is.
>
> so you can have a script that does an exit(1) and things calling that
> script can know that there was some sort of error in the script
> execution.
>
> (personally, i'd love to see exit() stop outputting the status,
> leaving that for die(), but whatever. i know better than to argue for
> consistency with perl. :)

Why not make exit() behaves like this:

exit(2): print nothing, set result status to 2
exit("error"); print "error", set no result.

I.e, on a numeric status, exit with that status, and if it's a string,
just print it. This way almost everybody gets what they want, AND it would
not break much scripts either.

Derick

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